We have ---- > 55, 62, 61, 54, 68, 72, 59 ,61, 70
In total 9 numbers
step 1
order the numbers from smallest to largest
54,55,59,61,61,62,68,70,72
1) What is the minimum of the data set ?
The lowest number is the minimum------------ > 54
2) What is the median of the data set ?
Find the number that is in the middle
54,55,59,61,61,62,68,70,72
The middle number is 61. That is the median.
Find the middle of each half. Since these numbers are between 2 numbers, you average the numbers (add then divide by 2).
55+59=114/2=57 This is the first quartile.
68+70=138/2=69 This is the third quartile
3) What is the first quartile of the data set ?
the answer is 57
4) What is the third quartile of the data set ?
the answer is 69
5) What is the maximum of the data set ?
The highest is the maximum------------ > 72
Answer:
Minimum = 54
Maximum = 72
Median = 61
First quartile = 57
Third quartile = 69
Step-by-step explanation:
Data : 55 62 61 54 68 72 59 61 70
Arrange the data is ascending order
54 , 55, 59, 61,61 ,62 , 68,70 ,72
So, minimum = 54
Maximum = 72
Median is the mid value .
There are nine observations.
So, 5th observation is the median
So, Median = 61
First quartile = It is the median of data of the left side data of the median .
Left side data : 54 , 55, 59, 61
There are four observations
Median = [tex]\frac{\frac{n}{2}th+(\frac{n}{2}+1)th}{2}[/tex]
Median = [tex]\frac{\frac{4}{2}th+(\frac{4}{2}+1)th}{2}[/tex]
Median = [tex]\frac{2th+3rd}{2}[/tex]
Median = [tex]\frac{55+59}{2}[/tex]
Median = [tex]57[/tex]
First quartile = 57
Third quartile = It is the median of data of the right side data of the median .
Right side data : 62 , 68,70 ,72
There are four observations
Median = [tex]\frac{\frac{n}{2}th+(\frac{n}{2}+1)th}{2}[/tex]
Median = [tex]\frac{\frac{4}{2}th+(\frac{4}{2}+1)th}{2}[/tex]
Median = [tex]\frac{2th+3rd}{2}[/tex]
Median = [tex]\frac{68+70}{2}[/tex]
Median = [tex]69[/tex]
Third quartile = 69
Hence minimum is 54 , maximum is 72 ,median is 61 , first quartile is 57 and third quartile is 69
